@wolfgang59
I would try to explain my logic thus....
Shown on previous page, that if a rectangle is a x b, then for the numerical size of area and perimeter to be equal, b = 2a/(a-2)
Divide out the RHS, we get b = 2 + 4/(a-2)
You have imposed the condition that a AND b must both be integers. Let a take any integer value. Can we discern when b is also integer?
The '2' is irrelevant, as clearly always an integer. Concentrate on the 4/(a -2)
If a is 7 or more, this is a proper fraction, always less than 1 (it simply gets smaller and smaller as a increases). It will never give an integer when added to 2.
So we need to look at a = 1 to 6 inclusive
If a = 1, b= -2 (impossible rectangle)
If a = 2, bottom line = 0, cannot divide by zero, impossible
If a = 3, b = 6 (tick)
If a = 4, b = 4, a solution, but a square
If a = 5, b = 10/3, not integer
If a = 6, b = 3 (a repeat of the only viable solution). qed